How do you enter the command for a cube root?
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I'm re-working the volume of a sphere equation (V=(4*pi*r^3)/3) to solve for the radius(r).
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John D'Errico
2016-9-7
编辑：John D'Errico
2016-9-7
Two simple options:
x^(1/3)
Or,
nthroot(x,3)
Be very careful though. If x is negative, it will return a complex number, because there are indeed THREE cube roots of a negative number. Two of them are complex. nthroot will give you the root you would expect however.
(-2)^(1/3)
ans =
0.62996 + 1.0911i
nthroot(-2,3)
ans =
-1.2599
In your case, it is not relevant, since the number will be non-negative.
2 个评论
James Tursa
2018-11-7
编辑：James Tursa
2018-11-7
"... there are indeed THREE cube roots of a negative number ..."
To complete John's thought, there are three distinct cube roots of every non-zero number (positive real, negative real, complex), not just of the negative real numbers. And as John points out, some of these roots are complex, so you need to know how the tools you are using behave in order to get the answer(s) you want. (In general, there are n distinct n'th roots of every non-zero real or complex number)
John D'Errico
2020-11-17
Good completion/correction. My statement was sloppy as I wrote it.
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